Question

5. Three resistors of resistances 5 ohm, 3 ohm and 2 ohm are connected in series with 10 V battery.
Calculate their effective resistance and the current flowing through the circuit.​

Answer 1

Answer:

your answers are:---

Effective resistance = 10 Ω

Current through the circuit = 1 A

given three resistances in series

R₁ = 5 Ω

R₂ = 3 Ω

R₃ = 2 Ω

Resistance in series = R₁ + R₂ + R₃

R = 5+3+2

R = 10 Ω

Ohm's Law   V= I R

V= applied voltage

I = current

R = resistance

as we all know In series current will be same through all the elements

I =v/r

I = 10/10

I = 1 A

Explanation:

Answer 2

Answer:

R1=5Ω

R2=2Ω

R3=3Ω

V= 10 V

As the resistors are connected in series, effective resistance= R1+R2+R3

Let the effective resistance be Re= 5+2+3=

Re= 10 Ω

V=IR

10= I(10)

I=10/10

I= 1 A